Consider a slowly-varying function $L:(1,\infty) \mapsto (0,\infty)$, i.e. a function such that $L(cx)/L(x)\to1$ as $x\to\infty$ for all $c>0$. Assume that $\lim_{x \to \infty}L(x)=0$.

Question: is it true that, necessarily, for any sequences $x_n \to \infty$ and $c_n \to \infty$, $L(x_nc_n)/L(x_n)=O(1)$?

Comments: To prove this is true, I was trying to use the following well known representation for slowly varying functions $$ L(x)=c(x) \exp\left( \int_{x_0}^x \frac{\epsilon(s)}{s} ds \right) $$ where $c:(1, \infty) \mapsto (0, \infty)$ satisfies $\lim_{x \to \infty}|c(x)| \in(0,\infty)$ and $\epsilon:(1,\infty) \mapsto \mathbb{R} $ is a continous bounded function on $[x_0,\infty)$, for some $x_0>0$. Unfortunately, this representation does not guarantee that, e.g., $L(x)$ is asymptotically equivalent to a nonincreasing function, which would have guaranteed $L(x_nc_n)/L(x_n) \leq 1$ (see, e.g., Remark B.1.11 in de Haan and Ferreira (2006) Extreme Value Theory: An Introduction). So I got stuck, I'm not sure whether I'm missing anything stupid.